Sin (2x) - cos (6x) = 0

Решение

Sin (2x) - cos (6x) = 0

cos (π/2 - 2x) - cos6x = 0

- 2sin[ (π/2 - 2x) + 6x]/2 * sin[ (π/2 - 2x) - 6x]/2 = 0

sin[ (π/2 + 4x) ]/2 * sin[ (π/2 - 8x) ]/2 = 0

1) sin[ (π/2 + 4x) ]/2 = 0

sin (π/4 + 2x) = 0

π/4 + 2x = πk, k ∈ Z

2x = - π/4 + πk, k ∈ Z

x = - π/8 + π/2, k ∈ Z

2) sin[ (π/2 - 8x) ]/2 = 0

sin (π/4 - 4x) = 0

sin (4x - π/4) = 0

4x - π/4 = πn, n ∈ Z

4x = π/4 + πn, n ∈ Z

x = π/16 + πn/4, n ∈ Z